Silberberg7e solution manual ch 19

CHAPTER 19 IONIC EQUILIBRIA IN
AQUEOUS SYSTEMS
FOLLOW–UP PROBLEMS
19.1A

Plan: The problems are both equilibria with the initial concentration of reactant and product given. For part (a), set
up a reaction table for the dissociation of HF. Set up an equilibrium expression and solve for [H3O+], assuming
that the change in [HF] and [F] is negligible. Check this assumption after finding [H3O+]. Convert [H3O+] to pH.
For part (b), first find the concentration of OH added. Then, use the neutralization reaction to find the change in
initial [HF] and [F]. Repeat the solution in part (a) to find pH.
a) Solution:
Concentration (M)
HF(aq)
+ H2O(l)

F(aq) +
H3O+(aq)
Initial
0.50
—
0.45
0
Change
–x
—
+x
+x
Equilibrium
0.50 – x
—
0.45 + x

x
Assumed that x is negligible with respect to 0.50 M and 0.45 M.
 H 3 O +   F 
 
Ka = 
 HF
[H3O+] = K a

 HF


F
 



= 6.8 x104

 0.45 = 7.5556x10
0.50

Check assumption: Percent error in assuming x is negligible:

–4

= 7.6x10–4

7.5556x104
100 = 0.17%.
0.45

The error is less than 5%, so the assumption is valid.
Solve for pH:
pH = –log (7.5556 x 10–4) = 3.12173 = 3.12
The other method to calculate the pH of a buffer is to use the Henderson-Hasselbalch equation:
[base]
pH = pKa + log
[acid]

[F ]
pKa = –log Ka = –log(6.8x10–4) = 3.16749
[HF]
[0.45]
pH = 3.16749 + log
[0.50]
pH = 3.12173 = 3.12
Since [HF] and [F] are similar, the pH should be close to pKa, which equals log (6.8x10–4) = 3.17. The pH should
be slightly less (more acidic) than pKa because [HF] > [F–]. The calculated pH of 3.12 is slightly less than pKa of
3.17.
b) Solution:
What is the initial molarity of the OH– ion?

 0.40 g NaOH  1 mol NaOH  1 mol OH 

 = 0.010 M OH

Molarity = 
L

 40.00 g NaOH  1 mol NaOH 

pH = pKa + log

Set up reaction table for neutralization of 0.010 M OH (note the quantity of water is irrelevant).
Concentration (M)
HF(aq)
+
OH(aq)  F(aq)
+
H2O(l)
Before addition
0.50
—
0.45
—
Addition
—
0.010
—
—
Change
– 0.010
– 0.010
+ 0.010________________
After addition
0.49
0
0.46
—
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19-1


Following the same solution path with the same assumptions as part (a):
 HF = 6.8 x104 0.49 = 7.2434782x10–4 = 7.2x10–4 M
[H3O+] = K a
 F 
0.46
 
Check assumption:





7.2434782x104
100 = 0.16%, which is less than the 5% maximum so the assumption is valid.
0.46
Solve for pH:
pH = –log (7.2434782x10–4) = 3.14005 = 3.14
or using the Henderson-Hasselbalch equation:
[F ]
[0.46]
pH = pKa + log
= 3.16749 + log
[HF]
[0.49]
pH = 3.14005 = 3.14
With addition of base, the pH should increase and it does, from 3.12 to 3.14. However, the pH should still be

slightly less than pKa: 3.14 is still less than 3.17.
19.1B

Plan: The problems are both equilibria with the initial concentration of reactant and product given. Take the
inverse log of –pKb to solve for the Kb. Then use Kb to solve for Ka: Ka x Kb = 1.0x10-14. For part (a), set up a
reaction table for the dissociation of (CH3)2NH2+ (the acid component of the buffer; Cl– is a spectator ion and does
not participate in the buffer reaction). Set up an equilibrium expression and solve for [H3O+], assuming that the
change in [(CH3)2NH2+] and [(CH3)2NH] is negligible. Check this assumption after finding [H3O+]. Convert
[H3O+] to pH. For part (b), first find the concentration of H3O+ added. Then, use the neutralization reaction to find
the change in initial [(CH3)2NH2+] and [(CH3)2NH]. Repeat the solution in part (a) to find pH.
a) Solution:
Kb = 10pKb = 103.23 = 5.8884x10–4 = 5.9x10–4
Ka =

Kw
Kb

=

1.0x10–14
5.9x10–4

= 1.7x10–11

Concentration (M) (CH3)2NH2+(aq)
+ H2O(l)
(CH3)2NH(aq)
Initial
0.25
—

0.30
Change
–x
—
+x
Equilibrium
0.25 – x
—
0.30 + x
Assumed that x is negligible with respect to 0.25 M and 0.30 M.
+
(CH3 ) 2 NH   H3O 
Ka =
(CH3 ) 2 NH 2+ 



+

H3O+(aq)
0
+x
x

 (CH3 ) 2 NH 2  
 = (1.7x10–11) (0.25) = 1.4167x10–11 = 1.4x10–11
[H3O+] = K a 
.
(CH 3 )2 NH 

Check assumption: Percent error in assuming x is negligible:

1.4x10–11
0.25

(100) = 5.6x10–9%

The error is less than 5%, so the assumption is valid.
Solve for pH:
pH = –log (1.4x10–11) = 10.8539 = 10.85
The other method to calculate the pH of a buffer is to use the Henderson-Hasselbalch equation:
[base]
pH = pKa + log
[acid]
(CH 3 ) 2 NH 
pH = pKa + log 
 (CH 3 ) 2 NH 2  



pKa = –log Ka = –log(1.7x10–11) = 10.7696

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19-2


pH = 10.7696 + log

(0.30)

pH = 10.8488 = 10.85

(0.25)

Since [(CH3)2NH2+] and [(CH3)2NH] are similar, the pH should be close to pKa, which equals
log(1.7x10–11) = 10.7696. The pH should be slightly greater (more basic) than the pKa because
[(CH3)2NH] > [(CH3)2NH2+]. The calculated pH of 10.85 is slightly greater than the pKa of 10.77.
b) Solution:
What is the initial molarity of the H3O+ ion?
Molarity =

0.73 g HCl

1 mol HCl

1 mol H3 O+

1L

36.46 g HCl

1 mol HCl

= 0.020 M H3O+

Set up reaction table for neutralization of 0.020 M H3O+ (note the quantity of water is irrelevant).
Concentration (M)
(CH3)2NH(aq)

+
H3O+ (aq)  (CH3)2NH2+ (aq)
+
H2O(l)
Before addition
0.30
—
0.25
—
Addition
—
0.020
—
—
Change
– 0.020
– 0.020
+ 0.020________________________
After addition
0.28
0
0.27
—
Following the same solution path with the same assumptions as part (a):
 (CH3 ) 2 NH 2  
 = (1.7x10–11) (0.27) = 1.6393x10–11 = 1.6x10–11
+
[H3O ] = K a 
.
(CH 3 )2 NH 

Check assumption: Percent error in assuming x is negligible:
The error is less than 5%, so the assumption is valid.
Solve for pH:
pH = –log (1.6x10–11) = 10.7959 = 10.80
or using the Henderson-Hasselbalch equation:
(CH 3 ) 2 NH 
pH = pKa + log 
 (CH 3 ) 2 NH 2  


pH = 10.7696 + log

(0.28)
(0.27)

1.6x10–11
0.25

(100) = 6.4x10–9%

pH = 10.7854 = 10.78

(The slight differences in the pH calculated using the reaction table and the pH calculated from the HendersonHasselbalch equation are due to differences in rounding.)
With addition of acid, the pH should decrease and it does, from 10.85 to 10.80.
19.2A

Plan: For high buffer capacity, the components of a buffer should be concentrated and the concentrations of the
base and acid components should be similar.
Solution:
a) The buffer has a much larger amount of weak acid than of the conjugate weak base. Addition of strong base

would convert some HB into B to make the ration [B]/[HB] closer to 1 according to the reaction:
+
H2O(l)
HB(aq)
+
OH(aq)  B(aq)
b) The buffer with the highest possible buffer capacity would have 4 HB particles and 4 B particles. Addition of
strong base would convert 3 HB particles to 3 B particles so that there are 4 of each of the weak acid and weak
base.

19.2B

Plan: The pH of a given buffer solution depends on the relative concentrations of conjugate base and acid. If
[A] > [HA], the pH of the buffer solution will be greater than its pKa. If [A] < [HA], the pH of the buffer
solution will be less than its pKa. Buffers are able to lessen the effects of added strong acid or base by neutralizing

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19-3


the strong acid or base. Added strong base reacts with the acid component of the buffer, and added strong acid
reacts with the conjugate base component of the buffer. The most effective buffers consist of a weak acid with a
pKa close to (within one pH unit of) the buffering pH and its conjugate base.
Solution:
a) The buffer with pH > pKa will have more conjugate base particles than weak acid particles. The buffer in scene
3 has 6 conjugate base particles to 3 weak acid particles and, thus, would have pH > pKa.
b) Strong base reacts with the weak acid component of the buffer, so the more weak acid there is, the more strong
base the buffer can react with. There are more weak acid particles in the buffer in scene 2 than in the other scenes.

c) Buffers are most effective at buffering pH values within one unit of their pKa. The pKa of this buffer, 4.2, is
more than one unit away from the buffering pH, 6.1, so this buffer would not be effective at buffering a pH of
6.1.
19.3A

Plan: Sodium benzoate is a salt so it dissolves in water to form Na+ ions and C6H5COO ions. Only the benzoate
ion is involved in the buffer system represented by the equilibrium:
C6H5COOH(aq) + H2O(l)

C6H5COO(aq) + H3O+(aq)
Given in the problem are the volume and pH of the buffer and the concentration of the base, benzoate ion. The
question asks for the mass of benzoic acid to add to the sodium benzoate solution. First, find the concentration of
C6H5COOH needed to make a buffer with a pH of 4.25. Multiply the volume by the concentration to find moles of
C6H5COOH and use the molar mass to find grams of benzoic acid.
Solution:
The concentration of benzoic acid is calculated from the Henderson-Hasselbalch equation:
[C H COO ]
pH = pK a + log 6 5
[C6 H5 COOH]
pKa = –log Ka = –log 6.3x10–5 = 4.20066
[0.050]
4.25 = 4.20066 + log
[x]
[0.050]
Raise each side to 10x .
0.04934 = log
[x]
[0.050]
1.1203146 =
[x]

x = 4.46303x10–2 M
The number of moles of benzoic acid is determined from the concentration and volume:
(4.46303x10–2 M C6H5COOH) x (5.0 L) = 0.2231515 mol C6H5COOH
Mass (g) of C6H5COOH =
 122.12 g C6 H5COOH 
 0.2231515 mol C6 H5COOH  
 = 27.2513 = 27 g C6H5COOH
 1 mol C6 H5COOH 
Prepare a benzoic acid/benzoate buffer by dissolving 27 g of C6H5COOH into 5.0 L of 0.050 M C6H5COONa.
Using a pH meter, adjust the pH to 4.25 with strong acid or base.

19.3B

Plan: Ammonium chloride is a salt so it dissolves in water to form NH4+ ions and Cl ions. Only the ammonium
ion is involved in the buffer system represented by the equilibrium:
NH4+(aq) + H2O(l)

NH3(aq) + H3O+(aq)
Given in the problem are the volume and pH of the buffer and the concentration of the conjugate base, ammonia.
The question asks for the mass of ammonium chloride to add to the ammonia solution. First, find the ratio of
ammonia to ammonium ion needed to make a buffer with a pH of 9.18. Multiply the volume of the solution by the
concentration of the ammonia to find moles of ammonia in the buffer. Then use the ratio of ammonia to
ammonium as a conversion factor to calculate the moles of ammonium ion. Then use the molar mass to find
grams of ammonium chloride.
Solution:
The ratio of ammonia to ammonium ion is calculated from the Henderson-Hasselbalch equation:

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19-4


pH = pKa + log

Ka of NH4+ =

Kw
Kb of NH3

 NH3 
 NH 4 



=

1.0x10–14
1.76x10–5

= 5.68x10–10

pKa = –log Ka = –log (5.68x10–10) = 9.25
 NH3 
9.18 = 9.25 + log
 NH 4 


–0.07 = log

0.85 =

 NH3 
 NH 4 



 NH3 
 NH 4 



Raise each side to 10x .

There are 0.85 moles of NH3 for every 1 mole of NH4+, or 0.85 mol NH3 = 1 mol NH4+.

Amount (mol) of NH3 in buffer = (0.75 L buffer)

0.15 mol NH3
1L

= 0.1125 mol NH3

 1 mol NH 4   1 mol NH 4 Cl   53.49 g NH 4 Cl 
Mass (g) of NH4Cl =  0.1125 mol NH3  
 = 7.1 g NH4Cl
 
 0.85 mol NH 



3  1 mol NH 4   1 mol NH 4 Cl 

19.4A

Plan: The titration is of a weak acid, HBrO, with a strong base, NaOH. The reactions involved are:
1) Neutralization of weak acid with strong base:
HBrO(aq) + OH(aq)  H2O(l) + BrO(aq)
Note that the reaction goes to completion and produces the conjugate base, BrO.
2) The weak acid and its conjugate base are in equilibrium based on the acid dissociation reaction:
a) HBrO(aq) + H2O(l)  BrO(aq) + H3O+(aq) or
b) the base dissociation reaction:
BrO–(aq) + H2O(l)  HBrO(aq) + OH–(aq)
The pH of the solution is controlled by the relative concentrations of HBrO and BrO.
For each step in the titration, first think about what is present initially in the solution. Then use the two reactions
to determine solution pH.
It is useful in a titration problem to first determine at what volume of titrant the equivalence point occurs. For part
(e), use the pH values from (a) – (d) to plot the titration curve.
Solution:
a) Before any base is added, the solution contains only HBrO and water. Equilibrium reaction 2a
applies and pH can be found in the same way as for a weak acid solution:
Concentration (M) HBrO(aq) + H2O(aq)  BrO(aq) + H3O+(aq)
Initial
0.2000
—
0
0
Change
–x
—
+x

+x
Equilibrium
0.2000 – x
—
x
x
 H 3 O    BrO  
 x  x 

 = 2.3x10–9 =
Ka = 
0.2000  x 
 HBrO
Assume 0.2000 – x = 0.2000 M
 x  x 
2.3x10–9 =
0.2000
x = [H3O+] = 2.144761x10–5 M
pH = –log [H3O+] = –log (2.144761x10–5) = 4.66862 = 4.67
Check the assumption by calculating the percent error in [HBrO]eq.

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19-5


2.144761x105
100 = 0.01%; this is well below the 5% maximum.
0.2000

b) When [HBrO] = [BrO], the solution contains significant concentrations of both the weak acid and its conjugate
base. Use the equilibrium expression for reaction 2a to find pH.
Since [HBrO] = [BrO], their ratio equals 1.
Percent error =

 H 3 O    BrO  


Ka = 
 HBrO

 HBrO 





= 2.3x109 1 = 2.3x10–9 M
 BrO  


pH = –log [H3O+] = –log (2.3x10–9) = 8.63827 = 8.64
Note that when [HBrO] = [BrO], the titration is at the midpoint (half the volume to the equivalence point) and
pH = pKa.
c) At the equivalence point, the total number of moles of HBrO present initially in solution equals the number of
moles of base added. Therefore, reaction 1 goes to completion to produce that number of moles of BrO. The
solution consists of BrO and water. Calculate the concentration of BrO, and then find the pH using the base
dissociation equilibrium, reaction 2b.
First, find equivalence point volume of NaOH.
3

 1 mol NaOH 
  1 mL 
1L
 0.2000 mol HBrO   10 L 
Volume (mL) of NaOH = 

  20.00 mL  

  3 

L
1
mL
1
mol
HBrO
0.1000
mol
NaOH




  10 L 

= 40.00 mL NaOH added
All of the HBrO present at the beginning of the titration is neutralized and converted to BrO at the equivalence point.
Calculate the concentration of BrO.
Initial moles of HBrO: (0.2000 M)(0.02000 L) = 0.004000 mol
Moles of added NaOH: (0.1000 M)(0.04000 L) = 0.004000 mol

Amount (mol)
HBrO(aq)
+
OH(aq)
 H2O(l) + BrO(aq)
Before addition
0.004000 mol
—
—
0
Addition
—
0.004000 mol
—
—
Change
– 0.004000 mol
– 0.004000 mol
—
+0.004000 mol
After addition
0
0
—
0.004000 mol
At the equivalence point, 40.00 mL of NaOH solution has been added (see calculation above) to make the total
volume of the solution (20.00 + 40.00) mL = 60.00 mL.
 0.004000 mol BrO    1 mL 
[BrO–] = 
  3  = 0.06666667 M


60.00 mL

  10 L 
Set up reaction table with reaction 2b, since only BrO and water are present initially:
BrO(aq) + H2O(l) 
HBrO(aq) +
OH(aq)
Concentration (M)
Initial
0.06666667
—
0
0
Change
–x
—
+x
+x
Equilibrium
0.06666667 – x
—
x
x
Kb = Kw/Ka = (1.0x10–14/2.3x10–9) = 4.347826x10–6
 OH    HBrO 
 x  x 

Kb = 
=

= 4.347826x10–6
0.06666667  x 
 BrO  



Assume that x is negligible, since [BrO] >> Kb.
 x  x 
4.347826x10–6 =
0.06666667
[H3O+] = K a

x = [OH] = 5.3838191x10–4 = 5.4 x 10–4 M
Check the assumption by calculating the percent error in [BrO]eq.
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19-6


5.3838191x104
100 = 0.8%, which is well below the 5% maximum.
0.06666667
pOH = –log (5.3838191x10–4) = 3.26891
pH = 14 – pOH = 14 – 3.26891 = 10.73109 = 10.73
d) After the equivalence point, the concentration of excess strong base determines the pH. Find the
concentration of excess base and use it to calculate the pH.
Initial moles of HBrO: (0.2000 M)(0.02000 L) = 0.004000 mol
Moles of added NaOH: 2 x 0.004000 mol = 0.008000 mol NaOH
+

OH(aq)
 H2O(l) + BrO(aq)
Amount (mol)
HBrO(aq)
Before addition
0.004000 mol
—
—
0
Addition
—
0.008000 mol
—
—
Change
– 0.004000 mol
– 0.004000 mol
—
+0.004000 mol
After addition
0
0.004000 mol
—
0.004000 mol
Excess NaOH: 0.004000 mol
1L

 1 mL 
Volume (mL) of added NaOH =  0.0080000 mol NaOH  



 0.1000 mol NaOH  103 L 
= 80.00 mL
Total volume: 20.00 mL + 80.00 mL = 100.0 mL
[NaOH] = 0.004000 mol NaOH/0.1000 L = 0.0400 M
pOH = –log (0.0400) = 1.3979
pH = 14 – pOH = 14 – 1.3979 = 12.60
e) Plot the pH values calculated in the preceding parts of this problem as a function of the volume of titrant.

Percent error =

Titration of HBrO with NaOH
14
13
12
11
10
pH

9
8
7
6
5
4
3
0

10 20 30 40 50 60 70 80 90 100
Volume of 0.1000 M NaOH, mL

The plot and pH values follow the pattern for a weak acid vs. strong base titration. The pH at the midpoint of the
titration does equal pKa. The equivalence point should be, and is, greater than 7.
19.4B

Plan: The titration is of a weak acid, C6H5COOH, with a strong base, NaOH. The reactions involved are:
1) Neutralization of weak acid with strong base:
C6H5COOH(aq) + OH(aq)  H2O(l) + C6H5COO(aq)
Note that the reaction goes to completion and produces the conjugate base, C6H5COO.

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19-7


2) The weak acid and its conjugate base are in equilibrium based on the acid dissociation reaction:
a) C6H5COOH(aq) + H2O(l)  C6H5COO(aq) + H3O+(aq) or
b) the base dissociation reaction:
C6H5COO–(aq) + H2O(l)  C6H5COOH(aq) + OH–(aq)
The pH of the solution is controlled by the relative concentrations of C6H5COOH and C6H5COO.
For each step in the titration, first think about what is present initially in the solution. Then use the two reactions
to determine solution pH.
Solution:
a) Before any base is added, the solution contains only C6H5COOH and water. Equilibrium reaction 2a
applies and pH can be found in the same way as for a weak acid solution:
Concentration (M) C6H5COOH(aq) +
Initial
0.1000
Change

–x
Equilibrium
0.1000 – x
C6 H5 HCOO– [H3 O+ ]

–5

Ka = 6.3x10 =

[C6 H5 COOH]

=

H2O(aq)
—
—
—

 C6H5COO(aq)
0
+x
x

+ H3O+(aq)
0
+x
x

(x)(x)
(0.10 – x)

Assume 0.1000 – x = 0.1000 M

Ka = 6.3x10–5 =

(x)(x)
(0.10)

x = [H3O+] = 2.50998x10–3 = 2.5x10–3M
pH = –log [H3O+] = –log (2.5x10–3) = 2.6021 = 2.60
Check the assumption by calculating the percent error in [C6H5COOH]eq.
2.5x103
0.10

(100) = 2.5% which is smaller than 5%, so the assumption is valid.

b) Any NaOH added to the buffer reacts with the benzoic acid. Calculate the amount (mol) of NaOH (OH)
added.
Amount (mol) of OH added = (12.00 mL OH)

1L

0.1500 mol OH–

1000 mL

1L

= 0.001800 mol OH

The added OH will react with the benzoic acid:
Initial moles of C6H5COOH: (0.1000 M)(0.03000 L) = 0.003000 mol
C6H5COOH(aq)
0.003000 mol
—
– 0.001800 mol
0.001200 mol

Amount (mol)
Before addition
Addition
Change
After addition

+

OH(aq)
 H2O(l) + C6H5COO(aq)
—
—
0
0.001800 mol
—
—
– 0.001800 mol
—
+0.001800 mol
0
—
0.001800 mol

At this point in the titration, 12.00 mL of NaOH solution has been added (see calculation above) to make the total
volume of the solution (12.00 + 30.00) mL = 42.00 mL (0.04200 L).
Calculate the new concentrations of [C6H5COOH] and [C6H5COO]:
Molarity of C6H5COOH =
Molarity of C6H5COO =

Ka = 6.3x10–5 =

0.001200 mol

0.04200 L
0.001800 mol
0.04200 L

C6 H5 HCOO– [H3 O+ ]

[H3O+] = (6.3x10–5)

[C6 H5 COOH]
(0.02857)
(0.04286)

= 0.02857 M
= 0.04286 M

, so [H3O+] = Ka

[C6 H5 COOH]
C6 H5 HCOO–

= 4.1995x10–5 = 4.2x10–5

pH = –log [H3O+] = –log (4.2x10–5) = 4.3768 = 4.38
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19-8


c) At the equivalence point, the total number of moles of C6H5COOH present initially in solution equals the
number of moles of base added. Therefore, reaction 1 goes to completion to produce that number of moles of
C6H5COO. The solution consists of C6H5COO and water. Calculate the concentration of C6H5COO, and then
find the pH using the base dissociation equilibrium, reaction 2b.
First, find equivalence point volume of NaOH.
Volume (mL) of NaOH =
(30.00 mL C6H5COOH)

1L

0.1000 mol C6 H5 COOH

1 mol NaOH

1L

1000 mL

1L

1 mol C6 H5 COOH

0.1500 mol NaOH

= 0.02000 L (20.00 mL) NaOH added
All of the C6H5COOH present at the beginning of the titration is neutralized and converted to C6H5COO at the
equivalence point. Calculate the concentration of C6H5COO.
Initial moles of C6H5COOH: (0.1000 M)(0.03000 L) = 0.003000 mol
Moles of added NaOH: (0.1500 M)(0.02000 L) = 0.003000 mol
C6H5COOH(aq)
0.003000 mol
—
– 0.003000 mol
0

Amount (mol)
Before addition
Addition
Change
After addition

+

OH(aq)
 H2O(l) + C6H5COO(aq)
—
—
0
0.003000 mol
—

—
– 0.003000 mol
—
+0.003000 mol
0
—
0.003000 mol

At this point in the titration, 20.00 mL of NaOH solution has been added (see calculation above) to make the total
volume of the solution (20.00 + 30.00) mL = 50.00 mL (0.05000 L).
Calculate the new concentration [C6H5COO]:
Molarity of C6H5COO =

0.003000 mol
0.05000 L

= 0.06000 M

Set up reaction table with reaction 2b, since only C6H5COO and water are present initially:
Concentration (M)
C6H5COO(aq) +
H2O(l) 
C6H5COOH (aq) +
Initial
0.06000
—
0
Change
–x
—

+x
Equilibrium
0.06000 – x
—
x
Kb = Kw/Ka = (1.0x10–14/6.3x10–5) = 1.6x10–10

Kb =

[C6 H5 COOH][OH ]
–

C6 H5 HCOO

=

[x][x]

0.06000 – x

OH(aq)
0
+x
x

= 1.6x10–10

Assume that x is negligible, since [C6H5COO] >> Kb.
1.6x10–10 =

[x][x]

0.06000

x = [OH] = 3.0984x10–6 = 3.1x10–6 M
Check the assumption by calculating the percent error in [C6H5COO]eq.
3.1x10–6
0.06000

(100) = 0.0052% which is smaller than 5%, so the assumption is valid.

pOH = –log (3.1x10–6) = 5.5086
pH = 14 – pOH = 14 – 5.50864 = 8.4914 = 8.49
d) After the equivalence point, the concentration of excess strong base determines the pH. Find the
concentration of excess base and use it to calculate the pH.
Initial moles of C6H5COOH: (0.1000 M)(0.03000 L) = 0.003000 mol
Moles of added NaOH: (0.1500 M)(0.02200 L) = 0.003300 mol
+
OH(aq)
 H2O(l) + C6H5COO(aq)
Amount (mol)
C6H5COOH(aq)
Before addition
0.003000 mol
—
—
0
Addition
—
0.003300 mol

—
—
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19-9


Change
– 0.003000 mol
– 0.003000 mol
—
+0.003000 mol
After addition
0
0.000300 mol
—
0.003000 mol
Excess NaOH: 0.000300 mol
Total volume: 22.00 mL + 30.00 mL = 52.00 mL (0.05200 L)
[NaOH] = 0.000300 mol NaOH/0.05200 L = 0.00577 M
pOH = –log (0.00577) = 2.2388
pH = 14 – pOH = 14 – 2.2388 = 11.76 (pH is shown to two decimal places)
19.5A

Plan: Write the formula of the salt and the reaction showing the equilibrium of a saturated solution. The ionproduct expression can be written from the stoichiometry of the solution reaction as the coefficients in the reaction
become exponents in the ion-product expression.
Solution:
a) The formula of calcium sulfate is CaSO4. The equilibrium reaction is:
CaSO4(s)  Ca2+(aq) + SO42(aq)

Ion–product expression: Ksp = [Ca2+][SO42]
b) Chromium(III) carbonate is Cr2(CO3)3.
Cr2(CO3)3(s)  2Cr3+(aq) + 3CO32(aq)
Ion–product expression: Ksp = [Cr3+]2[CO32]3
c) Magnesium hydroxide is Mg(OH)2.
Mg(OH)2(s)  Mg2+(aq) + 2OH(aq)
Ion–product expression: Ksp = [Mg2+][OH]2
d) Arsenic(III) sulfide is As2S3.
As2S3(s)  2As3+(aq) + 3S2(aq)
3{S2(aq) + H2O(l)  HS(aq) + OH(aq)}
As2S3(s) + 3H2O(l)  2As3+(aq) + 3HS(aq) + 3OH(aq)
The second equilibrium must be considered in this case because its equilibrium constant is large, so essentially all
the sulfide ion is converted to HS and OH.
Ion-product expression: Ksp = [As3+]2[HS]3[OH]3

19.5B

Plan: Examine the ion-product expressions. The exponents in the ion-product expression are the subscripts of
those ions in the chemical formula.
Solution:
a) The compound is lead(II) chromate. Its formula is PbCrO4.
b) The compound is iron(II) sulfide. Its formula is FeS. (Note: metal sulfides have a special form of the ionproduct expression. The sulfide ion, S2, is so basic that it reacts completely with water to form the hydrogen
sulfide ion, HS and hydroxide. Thus, the presence of the term [HS][OH] is an indication that the anion in the
compound is the sulfide ion, S2. Because the term [HS][OH] has an exponent of 1 in this ion-product
expression, there is one sulfide ion in the chemical formula for this compound.)
c) The compound is strontium fluoride. Its formula is SrF2.
d) The compound is copper(II) phosphate. Its formula is Cu3(PO4)2.

19.6A

Plan: Calculate the solubility of CaF2 as molarity and use molar ratios to find the molarity of
Ca2+ and F dissolved in solution. Calculate Ksp from [Ca2+] and [F] using the ion-product expression.
Solution: Convert the solubility to molar solubility:
 1.5 x104 g CaF2   1 mL   1 mol CaF2 
–4
Molarity = 
  3  
 = 1.9211x10 M CaF2

10.0
mL
78.08
g
CaF
10
L


2




[Ca2+] = [CaF2] = 1.9211x10–4 M because there is 1 mol of calcium ions in each mol of CaF2.
[F] = 2[CaF2] = 3.8422x10–4 M because there are 2 mol of fluoride ions in each mol of CaF2.
The solubility equilibrium is:
CaF2(s)  Ca2+(aq) + 2F(aq) Ksp = [Ca2+][F]2
Calculate Ksp using the solubility product expression from above and the saturated concentrations of calcium and
fluoride ions.
Ksp = [Ca2+][F]2 = (1.9211x10–4)( 3.8422x10–4)2 = 2.836024x10–11 = 2.8x10–11

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19-10


The Ksp for CaF2 is 2.8x10–11 at 18°C.
19.6B

Plan: Calculate the solubility of Ag3PO4 as molarity and use molar ratios to find the molarity of Ag+ and PO43
dissolved in solution. Calculate Ksp from [Ag+] and [PO43] using the ion-product expression.
Solution: Convert the solubility to molar solubility:
Molarity =

3.2x10–4 g Ag3 PO4

1000 mL

1 mol Ag3 PO4

50. mL

1L

418.7 g Ag3 PO4

= 1.5x10–5 M Ag3PO4

[Ag+] = 3[Ag3PO4] = 4.5x10–5 M because there are 3 mol of silver ions in each mol of Ag3PO4.

[PO43] = [Ag3PO4] = 1.5x10–5 M because there is 1 mol of phosphate ions in each mol of Ag3PO4.
The solubility equilibrium is:
 3 Ag+(aq) + PO43(aq)
Ksp = [Ag+]3[ PO43]
Ag3PO4 (s)
Calculate Ksp using the solubility product expression from above and the saturated concentrations of silver and
phosphate ions.
Ksp = [Ag+]3[ PO43] = (4.5x10–5)3(1.5x10–5) = 1.3669x10–18 = 1.4x10–18
The Ksp for Ag3PO4 is 1.4x10–18 at 20°C.
19.7A

Plan: Write the solubility reaction for Mg(OH)2 and set up a reaction table, where S is the unknown molar
solubility of the Mg2+ ion. Use the ion-product expression to solve for the concentration of Mg(OH)2 in a
saturated solution (also called the solubility of Mg(OH)2).
Solution:
Concentration (M)
Mg(OH)2(s)  Mg2+(aq) +
2OH(aq)
Initial
—
0
0
+2S
Change
—
+S
Equilibrium
—
S
2S

Ksp = [Mg2+][OH]2 = (S)(2S)2 = 4S3 = 6.3x10–10
S = 5.4004114x10–4 = 5.4x10–4 M Mg(OH)2
The solubility of Mg(OH)2 is equal to S, the concentration of magnesium ions at equilibrium, so the molar
solubility of magnesium hydroxide in pure water is 5.4x10–4 M.

19.7B

Plan: Write the solubility reaction for Ca3(PO4)2 and set up a reaction table, where S is the unknown molar
solubility of Ca3(PO4)2. Use the ion-product expression to solve for the concentration of Ca3(PO4)2 in a saturated
solution (also called the solubility of Ca3(PO4)2).
Solution:
Concentration (M)
Ca3(PO4)2(s)  3Ca2+(aq) + 2PO43(aq)
Initial
—
0
0
+2S
Change
—
+3S
Equilibrium
—
3S
2S
Ksp = [Ca2+]3[PO43]2 = (3S)3(2S)2 = 108S5 = 1.2x10–29
S = 6.4439x10–7 = 6.4x10–7 M Ca3(PO4)2
The solubility of Ca3(PO4)2 in pure water, S, is 6.4x10–7 M.

19.8A

Plan: Write the solubility reaction of BaSO4. For part (a) set up a reaction table in which
[Ba2+] = [SO42] = S, which also equals the solubility of BaSO4. Then, solve for S using the ion-product
expression. For part (b), there is an initial concentration of sulfate, so set up the reaction table including this initial
[SO42]. Solve for the solubility, S, which equals [Ba2+] at equilibrium.
Solution:
a) Set up reaction table.
Concentration (M)
BaSO4(s) 
Initial
—
Change
—
Equilibrium
—
Ksp = [Ba2+][SO42–] = S2 = 1.1x10–10
S = 1.0488x10–5 = 1.0x10–5 M

Ba2+(aq)
0
+S
S

+

SO42(aq)
0
+S
S

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19-11


The molar solubility of BaSO4 in pure water is 1.0x10–5 M.
b) Set up another reaction table with initial [SO42] = 0.10 M (from the Na2SO4).
BaSO4(s) 
Ba2+(aq) +
SO42(aq)
Concentration (M)
Initial
—
0
0.10
+S
Change
—
+S
Equilibrium
—
S
0.10 + S
Ksp = [Ba2+][SO42–] = S(0.10 + S) = 1.1x10–10
Assume that 0.10 + S is approximately equal to 0.10, which appears to be a good assumption based on the fact
that 0.10 > 1x10–10, Ksp.
Ksp = S(0.10) = 1.1x10–10
S = 1.1x10–9 M
Molar solubility of BaSO4 in 0.10 M Na2SO4 is 1.1x10–9 M.

The solubility of BaSO4 decreases when sulfate ions are already present in the solution. The calculated decrease is
from 10–5 M to 10–9 M, for a 10,000-fold decrease. This decrease is expected to be large because of the high
concentration of sulfate ions.
19.8B

Plan: Write the solubility reaction of CaF2. For part (a) set up a reaction table in which [Ca2+] = S, which also
equals the solubility of CaF2. Then, solve for S using the ion-product expression. For part (b), there is an initial
concentration of calcium, so set up the reaction table including this initial [Ca2+]. Solve for the solubility, S,
which equals [Ca2+] at equilibrium. For part (c), there is an initial concentration of fluoride, so set up the reaction
table including this initial [F]. Solve for the solubility, S, which equals [Ca2+] at equilibrium.
Solution:
a) Set up reaction table.
Concentration (M)
CaF2(s) 
Ca2+(aq) +
2F(aq)
Initial
—
0
0
+2S
Change
—
+S
Equilibrium
—
S
2S
Ksp = [Ca2+][ F]2 = (S)(2S)2 = 4S3 = 3.2x10–11
S = 2.0x10–4 M

The molar solubility of CaF2 in pure water is 2.0x10–4 M.
b) Set up another reaction table with initial [Ca2+] = 0.20 M (from the CaCl2).
Concentration (M)
CaF2(s) 
Ca2+(aq) +
2F(aq)
Initial
—
0.20
0
+2S
Change
—
+S
Equilibrium
—
0.20 + S
2S
Ksp = [Ca2+][ F]2 = (0.20 + S)(2S)2 = 1.1x10–10
Assume that 0.20 + S is approximately equal to 0.20, which appears to be a good assumption based on the fact
that 0.20 > 3.2x10–11, Ksp.
Ksp = (0.20)(2S)2 = 3.2x10–11
S = 6.3x10–6 M
Molar solubility of CaF2 in 0.20 M CaCl2 is 6.3x10–6 M.
b) Set up another reaction table with initial [F] = 0.40 M (from the NiF2; there are two fluoride ions per NiF2 unit,
so an NiF2 concentration of 0.20 M gives a fluoride ion concentration of 0.40 M.).
CaF2(s) 
Ca2+(aq) +
2F(aq)
Concentration (M)

Initial
—
0
0.40
+2S
Change
—
+S
Equilibrium
—
S
0.40 + 2S
Ksp = [Ca2+][ F]2 = (S)(0.40 + 2S)2 = 1.1x10–10
Assume that 0.40 + 2S is approximately equal to 0.40, which appears to be a good assumption based on the fact
that 0.40 > 3.2x10–11, Ksp.
Ksp = (S)(0.40)2 = 3.2x10–11
S = 2.0x10–10 M
Molar solubility of CaF2 in 0.20 M NiF2 is 2.0x10–10 M.
The solubility of CaF2 decreases when either calcium or fluoride ions are already present in the solution.

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19-12


19.9A

Plan: First, write the solubility reaction for the salt. Then, check the ions produced when the salt dissolves to see if
they will react with acid. Three cases are possible:

1) If OH is produced, then addition of acid will neutralize the hydroxide ions and shift the solubility equilibrium
toward the products. This causes more salt to dissolve. Write the solubility and neutralization reactions.
2) If the anion from the salt is the conjugate base of a weak acid, it will react with the added acid in a
neutralization reaction. Solubility of the salt increases as the anion is neutralized. Write the solubility and
neutralization reactions.
3) If the anion from the salt is the conjugate base of a strong acid, it does not react with a strong acid. The
solubility of the salt is unchanged by the addition of acid. Write the solubility reaction.
Solution:
a) Calcium fluoride, CaF2
Solubility reaction: CaF2(s)  Ca2+(aq) + 2F(aq)
Fluoride ion is the conjugate base of HF, a weak acid. Thus, it will react with H3O+ from the strong acid, HNO3.
Neutralization reaction: F(aq) + H3O+(aq)  HF(aq) + H2O(l)
The neutralization reaction decreases the concentration of fluoride ions, which causes the solubility equilibrium to
shift to the right and more CaF2 dissolves. The solubility of CaF2 increases with the addition of HNO3.
b) Zinc sulfide, ZnS
Solubility reaction: ZnS(s) + H2O(l)  Zn2+(aq) + HS(aq) + OH(aq)
Two anions are formed because the sulfide ion from ZnS reacts almost completely with water to form HS and
OH.
The hydroxide ion reacts with the added acid:
Neutralization reaction: OH(aq) + H3O+(aq)  2H2O(l)
In addition, the hydrogen sulfide ion, the conjugate base of the weak acid H2S, reacts with the added acid:
Neutralization reaction: HS(aq) + H3O+(aq)  H2S(aq) + H2O(l)
Both neutralization reactions decrease the concentration of products in the solubility equilibrium, which causes a
shift to the right, and more ZnS dissolves. The addition of HNO3 will increase the solubility of ZnS.
c) Silver iodide, AgI
Solubility reaction: AgI(s)  Ag+(aq) + I(aq)
The iodide ion is the conjugate base of a strong acid, HI. So, I will not react with added acid. The solubility of
AgI will not change with added HNO3.

19.9B

Plan: First, write the solubility reaction for the salt. Then, check the ions produced when the salt dissolves to see if
they will react with acid. Three cases are possible:
1) If OH is produced, then addition of acid will neutralize the hydroxide ions and shift the solubility equilibrium
toward the products. This causes more salt to dissolve. Write the solubility and neutralization reactions.
2) If the anion from the salt is the conjugate base of a weak acid, it will react with the added acid in a
neutralization reaction. Solubility of the salt increases as the anion is neutralized. Write the solubility and
neutralization reactions.
3) If the anion from the salt is the conjugate base of a strong acid, it does not react with a strong acid. The
solubility of the salt is unchanged by the addition of acid. Write the solubility reaction.
Solution:
a) Silver cyanide, AgCN
Solubility reaction: AgCN(s)  Ag+(aq) + CN(aq)
Cyanide ion is the conjugate base of HCN, a weak acid. Thus, it will react with H3O+ from the strong acid, HBr.
Neutralization reaction: CN(aq) + H3O+(aq)  HCN(aq) + H2O(l)
The neutralization reaction decreases the concentration of cyanide ions, which causes the solubility equilibrium to
shift to the right and more AgCN dissolves. The solubility of AgCN increases with the addition of HBr.
b) copper(I) chloride, CuCl
Solubility reaction: CuCl (s)  Cu+(aq) + Cl(aq)
The chloride ion is the conjugate base of a strong acid, HCl. So, Cl will not react with added acid. The solubility
of CuCl will not change with added HBr.
c) Magnesium phosphate, Mg3(PO4)2
Solubility reaction: Mg3(PO4)2(s)  Mg2+(aq) + 2PO43(aq)

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19-13

Phosphate ion is the conjugate base of HPO42, a weak acid. Thus, it will react with H3O+ from the strong acid,
HBr.
Neutralization reaction: PO43(aq) + H3O+(aq)  HPO42(aq) + H2O(l)
HPO42, in turn, is the conjugate base of H2PO4, a weak acid. Thus, it will react with H3O+ from the strong acid,
HBr.
Neutralization reaction: HPO42(aq) + H3O+(aq)  H2PO4(aq) + H2O(l)
H2PO4, in turn, is the conjugate base of H3PO4, a weak acid. Thus, it will react with H3O+ from the strong acid,
HBr.
Neutralization reaction: H2PO4(aq) + H3O+(aq)  H3PO4(aq) + H2O(l)
Each of these neutralization reactions ultimately decreases the concentration of phosphate ions, which causes the
solubility equilibrium to shift to the right and more Mg3(PO4)2 dissolves. The solubility of Mg3(PO4)2 increases
with the addition of HBr.
19.10A Plan: First, write the solubility equilibrium equation and ion-product expression. Use the given concentrations of
calcium and phosphate ions to calculate Qsp. Compare Qsp to Ksp.
If Ksp < Qsp, precipitation occurs. If Ksp  Qsp then, precipitation will not occur.
Solution:
Write the solubility equation:
Ca3(PO4)2(s)  3Ca2+(aq) + 2PO43(aq)
and ion-product expression: Qsp = [Ca2+]3[PO43]2
[Ca2+] = [PO43–] = 1.0x10–9 M
Qsp = [Ca2+]3[PO43]2 = (1.0x10–9)3(1.0x10–9)2 = 1.0x10–45
Compare Ksp and Qsp. Ksp = 1.2x10–29 > 1.0x10–45 = Qsp
Precipitation will not occur because concentrations are below the level of a saturated solution as shown by the
value of Qsp.
19.10B Plan: First, write the solubility equilibrium equation and ion-product expression. Find the concentrations of the
lead and sulfide ions in the final mixture. Use these concentrations to calculate Qsp. Compare Qsp to Ksp.
If Ksp < Qsp, precipitation occurs. If Ksp  Qsp then, precipitation will not occur.
Solution:
Write the solubility equation:
PbS(s)  Pb2+(aq) + S2(aq)

and ion-product expression: Qsp = [Pb2+][S2]
25 L of a solution containing Pb2+ are mixed with 0.500 L of a solution containing S2. The final volume is 25.5 L.
Amount (mol) of Pb2+ = (25 L)
Molarity of Pb2+ =

1 mol Pb2+

1L

207.2 g Pb2+

0.0018 mol Pb2+
25.5 L

Amount (mol) of S2: (0.500 L)
Molarity of S2 =

0.015 g Pb2+

0.050 mol S 2–
25.5 L

= 0.0018 mol Pb2+

= 7.1x10–5 M

0.10 mol S2–
1L

= 0.050 mol S2

= 2.0x10–3 M

Qsp = [Pb2+][S2] = (7.1x10–5)(2.0x10–3) = 1.4x10–7
Compare Ksp and Qsp. Ksp = 3x10–25 < 1.4x10–7 = Qsp
Precipitation will occur because concentrations are above the level of a saturated solution as shown by the value
of Qsp.
19.11A Plan: First, write the solubility equilibrium equation and ion-product expression. For b) use the given amounts of
nickel (II) and hydroxide ions to calculate Qsp. Compare Qsp to Ksp. For c) check the ions produced when the salt
dissolves to see if they will react with acid or if the hydroxide ion of a strong base is part of the ion-product
expression, in which case, it will influence the solubility of the solid through the common ion effect.
Solution:
a) Write the solubility equation:
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19-14


Ni(OH)2(s)  Ni2+(aq) + 2OH(aq)
Scene 3 has the same relative number of ions as in the formula of Ni(OH)2. The Ni2+ and OH ions are in a 1:2
ratio in Scene 3.
b) Write the ion-product expression: Qsp = [Ni2+][OH]2
Calculate Ksp using Scene 3: Ksp = [Ni2+][OH]2 = [2][4]2 = 32
Calculate Qsp using Scene 1: Qsp = [Ni2+][OH]2 = [3][4]2 = 48
Calculate Qsp using Scene 2: Qsp = [Ni2+][OH]2 = [4][2]2 = 16
Qsp exceeds Ksp in Scene 1 (48 > 32) so additional solid will form in Scene 1.
c) Hydroxide ion is one of the products of the solubility equilibrium reaction. The hydroxide ion reacts with
added acid:
Neutralization reaction: OH(aq) + H3O+(aq)  2H2O(l)

The neutralization reaction decreases the concentration of OH(aq) in the solubility equilibrium, which causes a
shift to the right, and more Ni(OH)2(s) dissolves. Addition of base (OH–) shifts the equilibrium to the left due to
the common-ion effect and the mass of Ni(OH)2 increases.
19.11B Plan: First, write the solubility equilibrium equation and ion-product expression. For b) use the given amounts of
lead(II) and chloride ions to calculate Qsp. Compare Qsp to Ksp. For c) check the ions produced when the salt
dissolves to see if they will react with acid or if the anion of the acid is part of the ion-product expression, in
which case, it will influence the solubility of the solid through the common ion effect.
Solution:
a) Write the solubility equation:
PbCl2(s)  Pb2+(aq) + 2Cl(aq)
Scene 1 has the same relative number of ions as in the formula of PbCl2. The Pb2+ and Cl ions are in a 1:2 ratio
in Scene 1.
b) Write the ion-product expression: Qsp = [Pb2+][Cl]2
Calculate Qsp using Scene 1: Qsp = [Pb2+][Cl]2 = [3][6]2 = 108
Calculate Qsp using Scene 2: Qsp = [Pb2+][Cl]2 = [4][5]2 = 100
Calculate Ksp using Scene 3: Ksp = [Pb2+][Cl]2 = [5][5]2 = 125
Qsp exceeds Ksp in Scene 3 (125 > 108) so additional solid will form in Scene 3.
c) Chloride ion is one of the products of the solubility equilibrium reaction. Added chloride ion (from HCl) will
shift the reaction to the left, due to the common ion effect. As a result of the reaction shifting to the left, the mass
of solid PbCl2 will increase.
19.12A Plan: Compare the Ksp values for the two salts. Since CaSO4 is more soluble, calculate the concentration of sulfate
ions in equilibrium with the Ca2+ concentration.
Solution:
The solubility equilibrium for CaSO4 is
CaSO4(s)  Ca2+(aq) + SO42(aq)
and Ksp = [Ca2+][SO42] = 2.4x105
Ksp
 2.4x105 
4
SO42  =

=

 [Ca 2+ ]  0.025 M  = 9.6x10 M


19.12B Plan: Compare the Ksp values for the two salts. Since BaF2 is more soluble, calculate the concentration of fluoride
ions in equilibrium with the Ba2+ concentration.
Solution:
The solubility equilibrium for BaF2 is
BaF2(s)  Ba2+(aq) + 2F(aq)
and Ksp = [Ba2+][F]2 = 1.5x106
[F] =

Ksp
2+

[Ba ]

=

1.5x10–6
(0.020)

= 8.7x103 M

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19-15

19.13A Plan: Write the complex-ion formation equilibrium reaction. Calculate the initial concentrations of Fe(H2O)63+ and
CN. The approach to complex ion equilibria problems is slightly different than for solubility equilibria because
formation constants are generally large while solubility product constants are generally very small. The best
mathematical approach is to assume that the equilibrium reaction goes to completion and then calculate back to
find the equilibrium concentrations of reactants. So, assume that all of the Fe(H2O)63+ reacts to form Fe(CN)63
and calculate the concentration of Fe(CN)63 formed from the given concentrations of Fe(H2O)63+ and CN and the
concentration of the excess reactant. Then use the complex ion formation equilibrium to find the equilibrium
concentration of Fe(H2O)63+.
Solution:
Equilibrium reaction: Fe(H2O)63+(aq) + 6CN(aq)

Initial concentrations from a simple dilution calculation:
Mf = MiVi/Vf
[Fe(H2O)63+] =
[CN] =

 3.1x10

2

M

  25.5 mL 

25.5 + 35.0 mL

Fe(CN)63(aq) + 6H2O(l)

= 0.0130661157 M

1.5 M  35.0 mL

= 0.867768595 M
25.5 + 35.0 mL
Set up a reaction table:
Concentration (M)
Fe(H2O)63+(aq) + 6CN(aq) Fe(CN)63(aq) +
Initial
0.0130661157
0.867768595
0
Change
0.0130661157 + x  6(0.013066115)
+0.013066115
Equilibrium
x
0.789371905
0.013066115
3

 Fe(CN)6 
0.0130661


=
Kf = 4.0x1043 =
6
 Fe(H 2 O)36    CN  
 x 0.7893719056





6H2O(l)
—
—
—

x = 1.35019x1045 = 1.4x1045 M
The concentration of Fe(H2O)63+ at equilibrium is 1.4x1045 M. This concentration is so low that it is impossible to
calculate it using the initial concentrations minus a variable x. The variable would have to be so close to the initial
concentration that the initial concentration of x cannot be calculated to enough significant figures (43 in this case)
to get a difference in concentrations of 1x1045 M. Thus, the approach above is the best to calculate the very low
equilibrium concentration of Fe(H2O)63+.
19.13B Plan: Write the complex-ion formation equilibrium reaction. Calculate the initial concentration of Al(H2O)63+. The
approach to complex ion equilibria problems is slightly different than for solubility equilibria because formation
constants are generally large while solubility product constants are generally very small. The best mathematical
approach is to assume that the equilibrium reaction goes to completion and then calculate back to find the
equilibrium concentrations of reactants. So, assume that all of the Al(H2O)63+ reacts to form AlF63 and calculate
the concentration of AlF63 formed from the given concentrations of Al(H2O)63+ and F. Then use the complex
ion formation equilibrium to find the equilibrium concentration of Al(H2O)63+.
Solution:
Equilibrium reaction: Al(H2O)63+(aq) + 6F(aq)

AlF63(aq) + 6H2O(l)
Initial concentrations:
3+

[Al(H2O)6 ] =



2.4 g AlCl3 x

1 mol AlCl3
133.33 g AlCl3

0.250 L

[F ] = 0.560 M
Set up a reaction table:
Al(H2O)63+(aq)
Concentration (M)
Initial
0.072
Change
0.072 + x
Equilibrium
x

= 0.072 M

+

6F(aq) AlF63(aq)
0.560
0
 6(0.072)
+0.072
0.128

0.072

+

6H2O(l)
—
—
—

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19-16


Kf = 4x1019 =

 AlF63 


3+  

 6

=

(0.072)
(x)(0.128)6

 Al(H 2 O)6


 F 
The concentration of Al(H2O)63+ at equilibrium is 4x1016 M.

x = 4.0927x1016 = 4x1016 M

19.14A Plan: Write equations for the solubility equilibrium and formation of the silver-ammonia complex ion. Add the
two equations to get the overall reaction. Set up a reaction table for the overall reaction with the given value for
initial [NH3]. Write equilibrium expressions from the overall balanced reaction and calculate Koverall from Kf and
Ksp values. Insert the equilibrium concentration values from the reaction table into the equilibrium expression and
calculate solubility.
Solution:
Equilibria:
Ksp = 5.0x10–13
AgBr(s)  Ag+(aq) + Br(aq)
+
+
Ag (aq) + 2NH3(aq)  Ag(NH3)2 (aq)
Kf = 1.7x107

+
AgBr(s) + 2NH3(aq)  Ag(NH3)2 (aq) + Br (aq) Koverall = KspKf
Set up reaction table:
AgBr(s)
+
2NH3(aq)  Ag(NH3)2+(aq) + Br(aq)
Concentration (M)
Initial
—
1.0

0
0
+S
+S
Change
—
– 2S
Equilibrium
—
1.0 – 2S
S
S


 Ag(NH3 )2   Br 

 = K K = (5.0x10–13)(1.7x107) = 8.5x10–6
Koverall = 
sp f
2
NH3 
Calculate the solubility of AgBr:
 S  S 
Koverall =
= 8.5x10–6
1.0  2S 2
Assume that 1.0 – 2S is approximately equal to 1.0, which appears to be a good assumption based on the fact that
1.0 >> 8.5x10–6, Koverall.
 S  S 
= 8.5x10–6

2
1.0
S = 2.9154759x10–3 = 2.9x10–3 M
The solubility of AgBr in ammonia is less than its solubility in hypo (sodium thiosulfate).
Since the formation constant for Ag(NH3)2+ is less than the formation constant of Ag(S2O3)23, the addition of
ammonia will increase the solubility of AgBr less than the addition of thiosulfate ion increases its solubility.

19.14B Plan: Write equations for the solubility equilibrium and formation of the lead(II)-hydroxide complex ion. Add the
two equations to get the overall reaction. Set up a reaction table for the overall reaction with the given value for
initial [OH]. Write equilibrium expressions from the overall balanced reaction and calculate Koverall from Kf and
Ksp values. Insert the equilibrium concentration values from the reaction table into the equilibrium expression and
calculate solubility.
Solution:
Equilibria:
PbCl2(s)  Pb2+(aq) + 2Cl(aq)
Ksp = 1.7x10–5
2+


Pb (aq) + 3OH (aq)  Pb(OH)3 (aq)
Kf = 8x1013



PbCl2(s)+ 3OH (aq)  Pb(OH)3 (aq) + 2Cl (aq) Koverall = KspKf
Set up reaction table:
PbCl2 (s)
+ 3OH (aq)  Pb(OH)3 (aq) + 2Cl
Concentration (M)
Initial

—
0.75
0
0
+S
+2S
Change
—
– 3S
Equilibrium
—
0.75 – 3S
S
2S
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19-17


 Pb(OH)3  Cl 



Koverall =
 OH   3



2

= KspKf = (1.7x10–5)(8x1013) = 1x109

Calculate the solubility of PbCl2:
S 2S 2

Koverall =

0.75 – 3S

4S3
0.75 – 3S
√
0.75 – 3S

3

3

= 1x109

= 1x109
Take the cube root of both sides of the equation

= 1x103

S = 472.4704 – 1889.8816S
S = 0.2499 = 0.25 M

CHEMICAL CONNECTIONS BOXED READING PROBLEMS

B19.1

Plan: Consult Figure 19.5 for the colors and pH ranges of the indicators.
Solution:
Litmus paper indicates the pH is below 7. The result from thymol blue, which turns yellow at a pH above 2.5,
indicates that the pH is above 2.5. Bromphenol blue is the best indicator as it is green in a fairly narrow range of
3.5 < pH < 4. Methyl red turns red below a pH of 4.3. Therefore, a reasonable estimate for the rainwater pH is
3.5 to 4.

B19.2

Plan: Find the volume of the rain received by multiplying the surface area of the lake by the depth of rain. Find
the volume of the lake before the rain. Express both volumes in liters. The pH of the rain is used to find the
molarity of H3O+; this molarity multiplied by the volume of rain gives the moles of H3O+. The moles of H3O+
divided by the volume of the lake plus rain gives the molarity of H3O+ and the pH of the lake.
Solution:
a) To find the volume of rain, multiply the surface area in square inches by the depth of rain. Convert the volume
in in3 to cm3 and then to L.
2

3

 4.840x103 yd2   36 in 
 2.54 cm 
Volume (L) of rain = 10.0 acres  
 
 1.00 in  

1 acre

 1 in 

  1 yd 
= 1.027902x106 L
+
At pH = 4.20, [H3O ] = 10–4.20 = 6.3095734x10–5 M
 6.3095734x10 5 mol 
Moles of H3O+ = 1.027902x106 L 
 = 64.8562 = 65 mol

L


b) Volume (L) of the lake =



 1 mL   103 L 


3 

 1 cm   1 mL 



2

3

 4.840x103 yd 2   36 in 
 12 in  2.54 cm   1 mL   103 L 
10.0 acres  
 

 10.0 ft  

 
3 

1 acre
 1 ft  1 in   1 cm   1 mL 

  1 yd 
= 1.23348x108 L
Total volume of lake after rain = 1.23348x108 L + 1.027902x106 L = 1.243759x108 L
mol H3O
64.8562 mol
=
= 5.214531x10–7 M
[H3O+] =
8
L
1.243759x10 L
pH = –log (5.214531x10–7) = 6.2827847 = 6.28
c) Each mol of H3O+ requires one mole of HCO3 for neutralization.
 1 mol HCO3
Mass (g) = (64.8562 mol H3O+) 
 1 mol H O 
3


= 3.97575x103 = 4.0x103 g HCO3–

  61.02 g HCO3
 

  1 mol HCO3





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19-18


END–OF–CHAPTER PROBLEMS

19.1

The purpose of an acid-base buffer is to maintain a relatively constant pH in a solution.

19.2

The weak-acid component neutralizes added base and the weak-base component neutralizes added acid so that the
pH of the buffer solution remains relatively constant. The components of a buffer do not neutralize one another
when they are a conjugate acid-base pair.

19.3

The presence of an ion in common between two solutes will cause any equilibrium involving either of them to
shift in accordance with Le Châtelier’s principle. For example, addition of NaF to a solution of HF will cause the
equilibrium
HF(aq) + H2O(l)  H3O+(aq) + F  (aq)
to shift to the left, away from the excess of F  , the co mmon ion .

19.4

a) Buffer 3 has equal, high concentrations of both HA and A . It has the highest buffering capacity.
b) All of the buffers have the same pH range. The practical buffer range is pH = pKa ± 1, and is independent
of concentration.
c) Buffer 2 has the greatest amount of weak base and can therefore neutralize the greatest amount of added acid.

19.5

A buffer with a high capacity has a great resistance to pH change. A high buffer capacity results when the weak
acid and weak base are both present at high concentration. Addition of 0.01 mol of HCl to a high capacity buffer
will cause a smaller change in pH than with a low capacity buffer, since the ratio [HA]/[A] will change less.

19.6

Only the concentration of the buffer components (c) has an affect on the buffer capacity. In theory, any
conjugate pair (of any pKa) can be used to make a high capacity buffer. With proper choice of components, it can
be at any pH. The buffer range changes along with the buffer capacity, but does not determine it. A high capacity
buffer will result when comparable quantities (i.e., buffer-component ratio < 10:1) of weak acid and weak base
are dissolved so that their concentrations are relatively high.

19.7

The buffer-component ratio refers to the ratio of concentrations of the acid and base that make up the buffer.
When this ratio is equal to 1, the buffer resists changes in pH with added acid to the same extent that it resists
changes in pH with added base. The buffer range extends equally in both the acidic and basic direction. When the
ratio shifts with higher [base] than [acid], the buffer is more effective at neutralizing added acid than base so the
range extends further in the acidic than basic direction. The opposite is true for a buffer where [acid] > [base].
Buffers with a ratio equal to 1 have the greatest buffer range. The more the buffer-component ratio deviates from
1, the smaller the buffer range.

19.8

pKa (formic acid) = 3.74; pKa (acetic acid) = 4.74. Formic acid would be the better choice, since its pKa is closer
to the desired pH of 3.5. If acetic acid were used, the buffer-component ratio would be far from 1:1 and the
buffer’s effectiveness would be lower. The NaOH serves to partially neutralize the acid and produce its conjugate
base.

19.9

Plan: Remember that the weak-acid buffer component neutralizes added base and the weak-base buffer
component neutralizes added acid.
Solution:
a) The buffer-component ratio and pH increase with added base. The OH reacts with HA to decrease its
concentration and increase [NaA]. The ratio [NaA]/[HA] thus increases. The pH of the buffer will be more basic
because the concentration of base, A, has increased and the concentration of acid, HA, decreased.
b) Buffer-component ratio and pH decrease with added acid. The H3O+ reacts with A to decrease its
concentration and increase [HA]. The ratio [NaA]/[HA] thus decreases. The pH of the buffer will be more acidic
because the concentration of base, A, has decreased and the concentration of acid, HA, increased.

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19-19


c) Buffer-component ratio and pH increase with the added sodium salt. The additional NaA increases the
concentration of both NaA and HA, but the relative increase in [NaA] is greater. Thus, the ratio increases and the
solution becomes more basic. Whenever base is added to a buffer, the pH always increases, but only slightly if the
amount of base is not too large.
d) Buffer-component ratio and pH decrease. The concentration of HA increases more than the concentration of
NaA, so the ratio is less and the solution is more acidic.
19.10

a) pH would increase by a small amount.
b) pH would decrease by a small amount.
c) pH would increase by a very small amount.
d) pH would increase by a large amount.

19.11

Plan: The buffer components are propanoic acid and propanoate ion, the concentrations of which are known.
The sodium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the propanoic aciddissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that
dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be
calculated from the Henderson-Hasselbalch equation.
Solution:
Concentration (M)
CH3CH2COOH(aq) + H2O(l)  CH3CH2COO(aq) + H3O+(aq)
Initial
0.15
—
0.35

0
Change
–x
—
+x
+x
Equilibrium
0.15 – x
—
0.35 + x
x
Assume that x is negligible with respect to both 0.15 and 0.35 since both concentrations are much larger than Ka.
 H3O  CH3CH 2 COO 


 =  x  0.35  x  =  x  0.35
–5
Ka = 1.3x10 =
0.15  x 
0.15
CH3CH 2 COOH 

CH3CH 2 COOH 
 0.15 
–6
–6
= 1.3x105 
 = 5.57143x10 = 5.6x10 M

CH3CH 2 COO 

 0.35 


Check assumption: Percent error = (5.6x10–6/0.15)100% = 0.0037%. The assumption is valid.
pH = –log [H3O+] = –log (5.57143x10–6) = 5.2540 = 5.25
Another solution path to find pH is using the Henderson-Hasselbalch equation:
 [base] 
pH = pK a + log 
pKa = –log (1.3x10–5) = 4.886

 [acid] 
x = [H3O+] = Ka =





 [CH 3 CH 2 COO  ] 
 [0.35] 
= 4.886 + log 
pH = 4.886 + log 

 [CH 3 CH 2 COOH] 
 [0.15] 


pH = 5.25398 = 5.25

19.12

C6H5COOH(aq) + H2O(l)  C6H5COO–(aq) + H3O+(aq)
 H3O  C6 H5 COO  


 = x 0.28  x  = x 0.28
Ka = 6.3x10–5 =
0.33  x
C
H
COOH
0.33
 6 5

+
–5
–5
–5
x = [H3O ] = (6.3x10 )(0.33/0.28) = 7.425x10 = 7.4x10 M
Check assumption: Percent error = (7.425x10–5/0.28)100% = 0.026%. The assumption is valid.
pH = –log [H3O+] = –log (7.425x10–5) = 4.1293 = 4.13

19.13

Plan: The buffer components are HNO2 and NO2, the concentrations of which are known. The potassium ion is a
spectator ion and is ignored because it is not involved in the buffer. Write the HNO2 acid-dissociation reaction and
its Ka expression. Set up a reaction table in which x equals the amount of acid that dissociates; solving for x will
result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be calculated from the
Henderson-Hasselbalch equation.
Solution:
Concentration (M)

HNO2(aq) + H2O(l) NO2(aq) + H3O+(aq)

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19-20


Initial
0.55
—
0.75
0
Change
x
—
+x
+x
Equilibrium
0.55  x
—
0.75 + x
x
Assume that x is negligible with respect to both 0.55 and 0.75 since both concentrations are much larger than Ka.
 H3O    NO 2 


 = x 0.75  x  = x 0.75
Ka = 7.1x10–4 =
0.55  x 

0.55
HNO2 
x = [H3O+] = Ka

HNO2 



= 7.1x104

 0.75
0.55

= 5.2066667x10–4 = 5.2x10–4 M
 NO2


Check assumption: Percent error = (5.2066667x10–4/0.55)100% = 0.095%. The assumption is valid.
pH = log [H3O+] = log (5.2066667x10–4) = 3.28344 = 3.28
Verify the pH using the Henderson-Hasselbalch equation.
 [base] 
pH = pK a + log 
pKa = –log(7.1x10–4) = 3.149

 [acid] 


 [NO 2  ] 
 [0.75] 
= 3.149 + log 

pH = 3.149 + log 

 [HNO ] 
 [0.55] 
2 

pH = 3.2837 = 3.28

19.14

HF(aq) + H2O(l)  F–(aq) + H3O+(aq)
 H 3 O    F 

   = x 0.25  x  = x 0.25
4
Ka = 6.8x10 =
0.20  x 
0.20
HF

HF

= (6.8x104)(0.20/0.25) = 5.44x104 = 5.4x104 M
 F 
 
Check assumption: Percent error = (5.44x104/0.20)100% = 0.27%. The assumption is valid.
pH = log [H3O+] = log (5.44x104) = 3.2644 = 3.26
Verify the pH using the Henderson-Hasselbalch equation.
[H3O+] = Ka

19.15

Plan: The buffer components are formic acid, HCOOH, and formate ion, HCOO, the concentrations of which are
known. The sodium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the
HCOOH acid-dissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of
acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the
pH can be calculated from the Henderson-Hasselbalch equation.
Solution:
Ka = 10  p K a = 103.74 = 1.8197x104
Concentration (M)
HCOOH(aq) + H2O(l)  HCOO(aq) + H3O+(aq)
Initial
0.45
—
0.63
0
Change
x
—
+x
+x
Equilibrium
0.45  x
—
0.63 + x
x
Assume that x is negligible because both concentrations are much larger than Ka.
 H3O   HCOO 



 = x 0.63  x  = x 0.63
4
Ka = 1.8197x10 =
0.45  x 
0.45
HCOOH

HCOOH



= 1.8197x104

 0.63
0.45

= 1.29979x104 = 1.3x104 M
 HCOO


Check assumption: Percent error = (1.29979x104/0.45)100% = 0.029%. The assumption is valid.
pH = log [H3O+] = log (1.29979x104) = 3.886127 = 3.89
Verify the pH using the Henderson-Hasselbalch equation.
x = [H3O+] = Ka



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19-21


 [base] 
pH = pK a + log 

 [acid] 
 [HCOO ] 
 [0.63] 
pH = 3.74 + log 
 = 3.74 + log 

[HCOOH]
 [0.45] 


pH = 3.8861 = 3.89
19.16

HBrO(aq) + H2O(l)  BrO(aq) + H3O+(aq)
Ka = 10  p K a = 108.64 = 2.2908677x109

 H3O   BrO 
x0.68  x x 0.68



Ka = 2.2908677x10 =
=
=

0.95  x
0.95
HBrO
9

HBrO

= (2.2908677x109)(0.95/0.68) = 3.2004769x109 = 3.2x109 M
 BrO 


Check assumption: Percent error = (3.2004769x109/0.68)100% = 0.00000047%. The assumption is valid.
pH = log [H3O+] = log (3.2004769x109) = 8.4947853 = 8.49
Verify the pH using the Henderson-Hasselbalch equation.
x = [H3O+] = Ka

19.17

Plan: The buffer components phenol, C6H5OH, and phenolate ion, C6H5O, the concentrations of which are
known. The sodium ion is a spectator ion and is ignored because it is not involved in the buffer. Write the
C6H5OH acid-dissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of
acid that dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the
pH can be calculated from the Henderson-Hasselbalch equation.
Solution:
Ka = 10  p K a = 10–10.00 = 1.0x10–10
Concentration (M)
C6H5OH(aq) + H2O(l)  C6H5O(aq) +
H3O+(aq)
Initial
1.2

—
1.3
0
Change
–x
—
+x
+x
Equilibrium
1.2 – x
—
1.3 + x
x
Assume that x is negligible with respect to both 1.0 and 1.2 because both concentrations are much larger than Ka.
 H 3 O    C6 H 5 O  


 = x 1.3  x  = x 1.3
Ka = 1.0x10–10 =
C6 H5OH 
1.2  x 
1.2
 1.2 
C6 H5OH 
–11
x = [H3O+] = Ka
= 1.0x1010 
 = 9.23077x10 M

 C6 H 5 O 

 1.3 


Check assumption: Percent error = (9.23077x10–11/1.2)100% = 7.7x10–9%. The assumption is valid.
pH = –log (9.23077x10–11) = 10.03476 = 10.03
Verify the pH using the Henderson-Hasselbalch equation:
 [base] 
pH = pK a + log 

 [acid] 





 [C H O  ] 
 [1.3] 
= 10.00 + log 
pH = 10.00 + log  6 5

 [C 6 H 5 OH] 
 [1.2] 


pH = 10.03

19.18

H3BO3(aq) + H2O(l)  H2BO3–(aq) + H3O+(aq)
Ka = 10  p K a = 10–9.24 = 5.7543994x10–10

 H3O   H 2 BO3 


 = x 0.82  x  = x 0.82
Ka = 5.7543994x10–10 =
H3 BO3 
0.12  x 
0.12

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19-22


H3BO3 
= (5.7543994x10–10)(0.12/0.82) = 8.4210723x10–11 M
 H2 BO3 


Check assumption: Percent error = (8.4210723x10–11/0.12)100% = 7.0x10–8%. The assumption is valid.
pH = –log [H3O+] = –log (8.4210723x10–11) = 10.0746326 = 10.07
Verify the pH using the Henderson-Hasselbalch equation.
x = [H3O+] = Ka

19.19

Plan: The buffer components ammonia, NH3, and ammonium ion, NH4+, the concentrations of which are known.
The chloride ion is a spectator ion and is ignored because it is not involved in the buffer. Write the NH4+ aciddissociation reaction and its Ka expression. Set up a reaction table in which x equals the amount of acid that
dissociates; solving for x will result in [H3O+], from which the pH can be calculated. Alternatively, the pH can be

calculated from the Henderson-Hasselbalch equation. The Ka of NH4+ will have to be calculated from the pKb.
Solution:
14 = pKa + pKb
pKa = 14 – pKb = 14 – 4.75 = 9.25
Ka = 10  p K a = 10–9.25 = 5.62341325x10–10
Concentration (M)
NH4+(aq) + H2O(l)  NH3(aq) + H3O+(aq)
Initial
0.15
—
0.25
0
Change
–x
—
+x
+x
Equilibrium
0.15 – x
—
0.25 + x
x
Assume that x is negligible with respect to both 0.25 and 0.15 because both concentrations are much larger than
Ka.

0.25  x H3O  0.25 H3O 
 NH 3   H 3O 
–10
Ka = .62341325x10 =
=

=
0.15  x 
 NH 4  

0.15


 NH4 
 0.15 

 =
–10
X = [H3O+] = Ka
5.62341325x10 10 
 = 3.374048x10 M
0.25
NH3 


Check assumption: Percent error = (3.374048x10–10/0.15)100% = 2x10–7%. The assumption is valid.
pH = –log [H3O+] = –log [3.374048x10–10] = 9.4718 = 9.47
Verify the pH using the Henderson-Hasselbalch equation.
 [base] 
pH = pK a + log 

 [acid] 





 [NH 3 ] 
 [0.25] 
= 9.25 + log 
pH = 9.25 + log 


+
 [NH ] 
 [0.15] 
4


pH = 9.47

19.20

Kb = 10  p K b = 10–3.35 = 4.4668359x10–4
The base component is CH3NH2 and the acid component is CH3NH3+. Neglect Cl–. Assume + x and – x are
negligible.
CH3NH2(aq) + H2O(l)
 CH3NH3+(aq) + OH–(aq)
CH3 NH3  OH  
0.60  x OH  0.60OH  




–4
Kb = 4.4668359x10 =

=
=
0.50  x 
0.50
CH3 NH 2 

CH 3 NH 2 
= (4.4668359x10–4)(0.50/0.60) = 3.7223632x10–4 M
CH3 NH 3 


Check assumption: Percent error = (3.7223632x10–4/0.50)100% = 0.074%. The assumption is valid.
pOH = –log [OH–] = –log (3.7223632x10–4) = 3.429181254
pH = 14.00 – pOH = 14.00 – 3.429181254 = 10.57081875 = 10.57
Verify the pH using the Henderson-Hasselbalch equation.
[OH–] = Kb

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19-23


19.21

Plan: The buffer components are HCO3 from the salt KHCO3 and CO32 from the salt K2CO3. Choose the Ka
value that corresponds to the equilibrium with these two components. The potassium ion is a spectator ion and is
ignored because it is not involved in the buffer. Write the acid-dissociation reaction and its Ka expression. Set up
a reaction table in which x equals the amount of acid that dissociates; solving for x will result in [H3O+], from
which the pH can be calculated. Alternatively, the pH can be calculated from the Henderson-Hasselbalch

equation.
Solution:
a) Ka1 refers to carbonic acid, H2CO3, losing one proton to produce HCO3. This is not the correct Ka because
H2CO3 is not involved in the buffer. Ka2 is the correct Ka to choose because it is the equilibrium constant for the
loss of the second proton to produce CO32 from HCO3.
b) Set up the reaction table and use Ka2 to calculate pH.
Concentration (M)
HCO3(aq) + H2O(l)  CO32(aq) + H3O+(aq)
Initial
0.22
—
0.37
0
Change
–x
—
+x
+x
Equilibrium
0.22 – x
—
0.37 + x
x
Assume that x is negligible with respect to both 0.22 and 0.37 because both concentrations are much larger than
Ka.
 H3O  CO32  

 = x 0.37  x  = x 0.37
Ka = 4.7x1011 = 
 HCO3 

0.22  x 
0.22


 HCO3 
 = 4.7x1011  0.22  = 2.79459x1011 M
[H3O+] = Ka 
 0.37 
2

CO3 




Check assumption: Percent error = (2.79459x1011/0.22)100% = 1.3x10-8%. The assumption is valid.
pH = –log [H3O+] = –log (2.79459x1011) = 10.5537 = 10.55
Verify the pH using the Henderson-Hasselbalch equation.
 [base] 
pH = pK a + log 
pKa = –log (4.7x10–11) = 10.328

 [acid] 





 [CO32  ] 
 [0.37] 

= 10.328 + log 
pH = 10.328 + log 

 [HCO  ] 
 [0.22] 
3


pH = 10.55

19.22

a) The conjugate acid-base pair is related by Ka2 (6.3x10–8).
b) Assume that x is negligible with respect to both 0.50 and 0.40 because both concentrations are much larger than
Ka. The acid component is H2PO4– and the base component is HPO42–. Neglect Na+. Assume + x and – x are
negligible.
H2PO4–(aq) + H2O(l)  HPO42–(aq) + H3O+(aq)
 H 3O    HPO 4 2  

 =  x  0.40  x  =  x  0.40
Ka = 6.3x10–8 = 
 H 2 PO 4  
0.50  x 
0.50


 H 2 PO4  
 = (6.3x10–8)(0.50/0.40) = 7.875x10–8 M
[H3O+] = Ka 
 HPO 4 2  



Check assumption: Percent error = (7.875x10–8/0.50)100% = 1.6x10–5%. The assumption is valid.
pH = –log [H3O+] = –log (7.875x10–8) = 7.103749438 = 7.10
Verify the pH using the Henderson-Hasselbalch equation.

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19-24


19.23

Plan: Given the pH and Ka of an acid, the buffer-component ratio can be calculated from the HendersonHasselbalch equation. Convert Ka to pKa.
Solution:
pKa = –log Ka = –log (1.3x10–5) = 4.8860566
 [base] 
pH = pK a + log 

 [acid] 

 [Pr  ] 
5.44 = 4.8860566 + log 

 [HPr] 
 [Pr  ] 
0.5539467 = log 

 [HPr] 

Raise each side to 10x.

[Pr  ]
= 3.5805 = 3.6
[HPr]
19.24

pKa = –log Ka = –log (7.1x10–4) = 3.14874165
 [base] 
pH = pK a + log 

 [acid] 
 [NO 2  ] 
2.95 = 3.14874165 + log 
 [HNO ] 
2 

 [NO 2  ] 
–0.19874165 = log 
 [HNO ] 
2 


Raise each side to 10x.

[NO 2  ]
= 0.632788 = 0.63
[HNO 2 ]
19.25

Plan: Given the pH and Ka of an acid, the buffer-component ratio can be calculated from the HendersonHasselbalch equation. Convert Ka to pKa.
Solution:
pKa = –log Ka = –log (2.3x10–9) = 8.63827
 [base] 
pH = pK a + log 

 [acid] 

 [BrO ] 
7.95 = 8.63827 + log 

 [HBrO] 
 [BrO ] 
–0.68827 = log 

 [HBrO] 

Raise each side to 10x.

[BrO ]
= 0.204989 = 0.20
[HBrO]
19.26

Given the pH and Ka of an acid, the buffer-component ratio can be calculated from the Henderson-Hasselbalch
equation.
pKa = –log Ka = –log (1.8x10–5) = 4.744727495
 [base] 
pH = pK a + log 


 [acid] 
 [CH 3 CO O  ] 
4.39 = 4.744727495 + log 
 [CH 3 COOH] 



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19-25